Chapter 17:  Acids and Bases

 

Section 17-1: Arrhenius Theory, pH, and pOH

Section 17-2: Monoprotic Strong Acids and Strong Bases

Section 17-3: Monoprotic Weak Acids

Section 17-4: Polyprotic Weak Acids and Sulfuric Acid

Section 17-5: Bronsted Theory

Section 17-6: Weak Bases

Section 17-7: Lewis Structures and Acid Strength

Section 17-8: Hydrolysis - Aqueous Ions as Acids and Bases

Section 17-9: Reactions of Nonmetal Oxides with Water to Create Oxoacids

Section 17-10: Experiment - Determining the Molar Mass of an Unknown Monoprotic Weak Acid by Titration

Section 17-11:  pH of Buffer Solutions

Section 17-12:  pH and Titration Curves

Section 17-13:  Acid-Base Indicators

Chapter 17 Practice Exercises and Review Quizzes

 

 

 

 

Section 17-1:  Arrhenius Theory, pH, and pOH

Arrhenius theory essentially states that an acid ionizes in water to produce H⁺, known as a hydrogen ion or a proton, while a base ionizes in water to produce OH^-, hydroxide ion.  One way to express the level of acidity or basicity in an aqueous solution is to indicate the molarity of H⁺, [H⁺], or the molarity of OH^-, [OH^-].  Alternatively, pH and pOH (both expressed without a unit) can be used:

 

Equation 1:  pH = - log [H⁺]

 

Equation 2:  pOH = - log [OH^-]

 

[Note:  When rounding the result y in a calculation y = - log x, the number of decimal places in y should equal the number of significant figures in x.]

 

In any aqueous solution, the equilibrium reaction known as the autoionization of water occurs:

 

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)

The equilibrium constant K_c for the reaction above is typically replaced by K_w in the equilibrium expression:

 

Equation 3:  K_w = [H⁺][OH^-]

 

At 25°C, K_w = 1.0 x 10^-14.  Therefore, we can derive an equation relating pH and pOH at 25°C as follows:

 

- log (K_w) = -log ([H⁺][OH^-])

- log (1.0 x 10^-14) = (- log [H⁺]) + (-log [OH^-])

Equation 4:  14.00 = pH + pOH

 

On the pH scale at 25°C:

 

pH < 7 = acidic

pH 7 = neutral

pH > 7 = basic

 

In this chapter, unless otherwise indicated, assume all solutions are aqueous and at 25°C.

 

 

 

 

Section 17-2:  Monoprotic Strong Acids and Strong Bases

If an acid molecule is only capable of releasing one H⁺ (has only one ionizable proton), the acid is said to be monoprotic.  We can assume that all molecules of the monoprotic strong acids HCl, HBr, HI, HNO₃, and HClO₄ will ionize completely in solution to produce one proton and an anion having a 1- charge, with the reverse reaction being negligible (and, thus, we will use a single forward arrow rather that a double equilibrium arrow in the equation).  Therefore, the initial molarity (M_i) of the strong acid will become the molarity of H⁺ present in the solution at equilibrium, as demonstrated below on the RICE chart for a nitric acid solution:

 

 

Once we know the molarity of H⁺ at equilibrium, we can also calculate the molarity of hydroxide ion (which, even in an acidic solution, is produced due to the autoionization of water), pH, and pOH as follows:

 

Sample Exercise 17A:

 

A 44 mL sample of HCl gas, measured at 45°C and 711 torr, was dissolved in water to yield 18.5 mL of solution.  Calculate the molarity of hydrogen ion, the molarity of hydroxide ion, pH, and pOH in this solution.

 

Solution:

 

First, use the Ideal Gas Law to calculate the moles of HCl dissolved, then divide by the volume of solution to determine the initial molarity of HCl:

 

 

 

Since HCl is a strong acid, we now complete a RICE chart showing complete ionization similar to that for HNO₃ above:

 

 

We can now complete the calculations using the equations above:

 

 

(Note that we could have also calculated pOH using Equation 2.)

 

Soluble metal hydroxides are strong bases that will ionize completely in water.  If there is only one hydroxide ion per formula unit, the initial molarity of the metal hydroxide will become the molarity of hydroxide ion present in the solution at equilibrium, as demonstrated below on the RICE chart for a sodium hydroxide solution:

 

 

However, if there are two hydroxide ions per formula unit, the molarity of hydroxide ion present in the solution at equilibrium will be twice the initial molarity of the metal hydroxide:

 

Sample Exercise 17B:

A 0.0070 gram sample of calcium hydroxide was dissolved in water to create 64 mL of solution.  Calculate the molarity of hydroxide ion, the molarity of hydrogen ion, pOH, and pH in this solution.

 

Solution:

 

First, convert the mass of calcium hydroxide to moles, then divide by the volume of solution to determine the initial molarity of calcium hydroxide:

 

 

Since calcium hydroxide is a strong base, we now complete a RICE chart showing complete ionization similar to that for NaOH above:

 

 

 

 

 

 

Section 17-3:  Monoprotic Weak Acids

 

For the remainder of this chapter, assume that any acid that is not included on the strong acid list of HCl, HBr, HI, HNO₃, or HClO₄ is a weak acid, meaning that only a fraction of the acid molecules ionize in water.  For the ionization of a monoprotic weak acid, the RICE chart and pH calculation will differ from that of a monoprotic strong acid in the following ways:

 

1. A double equilibrium arrow will be used in the ionization equation to indicate that the reverse reaction is significant.

 

2. The initial molarity of the weak acid will not decrease to zero as the reaction proceeds.  As such, the molarity of H⁺ present at equilibrium will be less than the initial molarity of the weak acid.

 

3. To determine the magnitude of the change (C) line on the RICE chart, we will need to do an equilibrium calculation.  This will require knowing the value of K_c for the ionization reaction, which is typically expressed as K_a, known as the acid ionization constant.

 

The RICE chart and equilibrium calculation for all monoprotic weak acids will follow the format below:

 

 

(Note that the anion A^- may contain more hydrogen atoms, but we can assume that these are not ionizable.)

 

 

 

Once we know the value of x, we can also calculate the percentage of the original sample of acid molecules that have been ionized as follows:

 

 

When comparing different monoprotic weak acids with the same initial molarity:

 

larger K_a = larger x = larger % ionization = larger [H⁺] = lower pH = more acidic = stronger acid

   

Sample Exercise 17C:

Write the acid ionization equation and calculate the pH and percent ionization of 0.16 M hydrofluoric acid, HF (K_a = 6.8 x 10^-4).

 

Solution:

 

We complete the RICE chart and equilibrium calculation for a monoprotic weak acid, then use the value of x = [H⁺] to calculate the pH and percent ionization:

 

 

 

 

 

If we know the pH of a solution, we can calculate [H⁺] as follows:

 

Equation 1a:  [H⁺] = 10^-pH 

 

If we know the initial molarity and the pH of a monoprotic weak acid solution, we can calculate K_a as follows:

 

 Sample Exercise 17D:

A 0.58 M acetic acid solution has a pH of 2.49.  Write the acid ionization equation and calculate K_a for acetic acid, HC₂H₃O₂ (a more condensed version of the formula CH₃COOH showing the one ionizable proton at the beginning, separate from the other non-ionizable hydrogens). 

 

Solution:

 

 

 

 

If we know the initial molarity and the percent ionization for a monoprotic weak acid, we can calculate pH and K_a, as demonstrated at the end of the chapter in Practice Exercise 17-5.  If we know the pH and K_a for a monoprotic weak acid solution, we can calculate the initial molarity, as demonstrated in Practice Exercise 17-6.

 

Relative strengths of weak acids can also be compared using pK_a values:

 

pK_a = -log K_a

K_a = 10^-pKa

lower pK_a = larger K_a = stronger acid

 

 

 

Section 17-4:  Polyprotic Weak Acids and Sulfuric Acid

A polyprotic weak acid has more than one ionizable proton and will ionize in a stepwise series of reactions, each of which has its own distinct K_a value. 

 

A diprotic weak acid will ionize in two steps as follows:

 

1^st ionization:  H₂A (aq) ⇌ H⁺ (aq) + HA^- (aq)     K_c = K_a1

 

2^nd ionization:  HA^- (aq) ⇌ H⁺ (aq) + A^2- (aq)     K_c = K_a2

 

A triprotic weak acid will ionize in three steps as follows:

 

1^st ionization:  H₃A (aq) ⇌ H⁺ (aq) + H₂A^- (aq)     K_c = K_a1

 

2^nd ionization:  H₂A^- (aq) ⇌ H⁺ (aq) + HA^2- (aq)     K_c = K_a2

 

3^rd ionization:  HA^2- (aq) ⇌ H⁺ (aq) + A^3- (aq)     K_c = K_a3

 

The K_a values will decrease with each successive ionization:  K_a1 > K_a2 > K_a3.  In general, only the 1^st ionization will produce enough H⁺ to affect the pH, so we will only use the RICE chart and equilibrium calculation from the 1^st ionization to calculate the pH:   

 

Sample Exercise 17E:

 

Write the stepwise acid ionization equations and calculate the pH of 1.6 M phosphoric acid, H₃PO₄, which has the following acid ionization constants:

 

K_a1 = 7.5 x 10^-3

K_a2 = 6.2 x 10^-8

K_a3 = 4.2 x 10^-13

 

Solution:

 

We obtain the pH by completing the RICE chart and equilibrium calculation for the 1^st ionization only, after which we proceed with writing the 2^nd and 3^rd ionization equations:

 

 

 

 

2^nd ionization:  H₂PO₄^- (aq) ⇌ H⁺ (aq) + HPO₄^2- (aq)    

 

3^rd ionization:  HPO₄^2- (aq) ⇌ H⁺ (aq) + PO₄^3- (aq)  

 

 

Sulfuric acid is a unique case where the 1^st ionization of H₂SO₄ occurs completely with a negligible reverse reaction (similar to a monoprotic strong acid), whereas only a fraction of the HSO₄^- anions further ionize in the 2^nd ionization (similar to a diprotic weak acid):

 

1^st ionization:  H₂SO₄ (aq) → H⁺ (aq) + HSO₄^- (aq)

 

2^nd ionization:  HSO₄^- (aq) ⇌ H⁺ (aq) + SO₄^2- (aq)    

 

 

Section 17-5:  Bronsted Theory

Because ammonia, NH₃, does not contain hydroxide ion, Arrhenius theory does not explain how ammonia can function as a base in water. On the other hand, Bronsted theory can explain how a solution of ammonia can be basic by defining an acid and base as follows:

 

Bronsted acid = proton (H⁺) donor

Bronsted base = proton (H⁺) acceptor

 

In the case of an ammonia solution, water will act as a Bronsted acid to donate a proton to the Bronsted base ammonia.  Since the Bronsted acid water loses a proton, the formula will decrease by one hydrogen and the charge will decrease by 1, thus producing the hydroxide ion that makes the solution basic.  At the same time, the Bronsted base ammonia will gain a proton, so the formula will increase by one hydrogen and the charge will increase by 1 to produce the ammonium ion:

 

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH^- (aq) 

 

Although we have used Arrhenius theory to write acid ionization equations and will continue to do so, we could have chosen to use Bronsted theory instead.  In the case of solution containing the monoprotic weak acid HA, water can act as a Bronsted base to accept a proton from the Bronsted acid HA:

 

HA (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^- (aq) 

 

The cation H₃O⁺ is known as the hydronium ion and can replace H⁺ when calculating pH after writing the acid ionization equation according to Bronsted theory:  pH = - log [H₃O⁺].  Since liquid water will be omitted from the RICE chart and K_a expression, the pH calculated using Bronsted theory will be the same as the pH calculated using Arrhenius theory.

 

As shown above, water acts as a Bronsted acid in an ammonia solution, but a Bronsted base in an acid solution.  A molecule or ion that can act as both a Bronsted acid or a Bronsted base in different situations is said to be amphoteric (or amphiprotic).

 

Bronsted theory also applies to proton transfer reactions in which water is not a reactant or a product:

 

Sample Exercise 17F:

 

Identify the Bronsted acids and bases in the forward and reverse directions for the reaction below:

 

CO₃^2- (aq) + HC₂O₄^- (aq) ⇌ HCO₃^- (aq) + C₂O₄^2- (aq) 

 

Solution:

 

forward reaction:  Bronsted acid = HC₂O₄^- (donates proton), Bronsted base = CO₃^2- (accepts proton)

reverse reaction:  Bronsted acid = HCO₃^-, Bronsted base = C₂O₄^2-

 

 

After a proton has been donated by a Bronsted acid, the resulting molecule or ion is known as the conjugate base.  After a proton has been accepted by a Bronsted base, the resulting molecule or ion is known as the conjugate acid:  

 

Sample Exercise 17G:

 

Write the formula for:

 

a. the conjugate acid of HS^-

b. the conjugate base of HC₄H₄O₆^-

 

Solution:

 

a. conjugate acid = add H⁺ to formula = H₂S

b. conjugate base = remove H⁺ from formula = C₄H₄O₆^2-

 

 

Section 17-6:  Weak Bases

 

For a solution of a neutral weak base B, the base ionization equation can be represented: 

 

B (aq) + H₂O (l) ⇌ BH⁺ (aq) + OH^- (aq)

(proton is typically added at end of base formula)

 

K_c for the reaction above can be expressed as K_b, known as the base ionization constant.  To calculate the pH and percent ionization in a weak base solution, we complete a RICE chart and equilibrium calculation similar to that for a monoprotic weak acid.  However, the column under liquid water will be omitted from the RICE chart and water will be omitted from the K_b expression.  Also note that the calculated value of x = [OH^-], not the molarity of hydrogen ions:

 

 

 

 

When comparing different weak bases with the same initial molarity:

 

larger K_b = larger x = larger % ionization = larger [OH^-] = higher pH = more basic = stronger base

 

Sample Exercise 17H:

 

Calculate the pH and percent ionization of 0.029 M ammonia (K_b = 1.8 x 10^-5).

 

Solution:

 

 

 

 

 

If we know the pOH of a solution, we can calculate [OH^-] as follows:

 

Equation 2a:  [OH^-] = 10^-pOH 

 

If we know the initial molarity and the pH of a weak base solution, we can calculate K_b as follows:

 

Sample Exercise 17I:

 

A 0.026 M methylamine solution has a pH of 11.51.  Write the base ionization equation and calculate K_b for methylamine, CH₃NH₂.

 

 

Solution:

 

 

 

 

 

If we know the initial molarity and the percent ionization for a weak base, we can calculate pH and K_b, as demonstrated in Practice Exercise 17-12.  If we know the pH and K_b for a weak base solution, we can calculate the initial molarity, as demonstrated in Practice Exercise 17-13.

 

Relative strengths of weak bases can also be compared using pK_b values:

 

pK_b = -log K_b

K_b = 10^-pKb

lower pK_b = larger K_b = stronger base

 

 

Section 17-7:  Lewis Structures and Acid Strength

 

The Lewis structure for an oxoacid (or oxyacid) with the formula H_mXO_n will generally show the center atom X bonded to each oxygen atom, and each hydrogen atom will be bonded to a different oxygen atom, as demonstrated below for sulfurous acid, H₂SO₃:

 

 

(other resonance structures also acceptable)

 

To compare the relative strengths of oxoacids, use the following guidelines:

 

1. More terminal oxygens (= oxygens not bonded to a hydrogen = n-m) = oxoacid donates proton more readily = stronger acid.

2. Higher electronegativity of center atom X = oxoacid donates proton more readily = stronger acid.

 

Sample Exercise 17J:

 

Draw Lewis structures for nitric acid, HNO₃, and phosphoric acid, H₃PO₄.  Which is the stronger acid?  Give two reasons to justify your answer.

 

Solution:

 

 

 

 

Nitric acid is stronger acid because:

 

1. HNO₃ has more terminal oxygens (3-1=2) than H₃PO₄ (4-3=1).

2. Electronegativity for central atom N is higher than for central atom P.

 

 

The Lewis structure for a monoprotic organic acid with the formula RCOOH, where R can either represent a single atom or a group of atoms, is shown below:

 

 

The ionizable proton will be the one single-bonded to the oxygen atom on the right end of the Lewis structure, not in R.  To compare the relative strengths of organic acids, use the following general guidelines:

 

1. Higher electronegativity of atoms in R = organic acid donates proton more readily = stronger acid.

2. Highly electronegative atoms in R are closer in proximity to ionizable hydrogen = organic acid donates proton more readily = stronger acid.

 

Sample Exercise 17K:

Which of the two acids shown below is the stronger acid?  Give two reasons to justify your answer.

 

 

 

 

Solution:

The bottom acid is stronger because:

 

1. Electronegativity of F is higher than Br.

2. F is closer to ionizable proton than Br.

 

Section 17-8:  Hydrolysis – Aqueous Ions as Acids and Bases

When an ionic compound dissolves in water, it is sometimes possible for the ions in solution to function as acids or bases in a process known as hydrolysis.  Use the following guidelines to decide if an ion will hydrolyze and, therefore, affect the pH:

 

1. A cation BH⁺ containing an ionizable proton (ammonium, NH₄⁺, for example) will generally hydrolyze as a weak acid:

 

BH⁺ (aq) ⇌ H⁺ (aq) + B (aq)

 

We can determine K_a for the reaction above by utilizing the fact that when two reactions are added to obtain a third reaction, the product of the equilibrium constants for the two added reactions will equal the equilibrium constant for the third reaction:

 

BH⁺ (aq) ⇌ H⁺ (aq) + B (aq)     K_a for BH⁺ = ???

B (aq) + H₂O (l) ⇌ BH⁺ (aq) + OH^- (aq)     K_b for B = given in problem or found on K_b data table

__________________________________________________________________________________________

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)     K_w = 1.0 x 10^-14

 

(K_a for BH⁺)(K_b for B) = K_w

 

 

(so stronger base B = weaker conjugate acid BH⁺)

 

2. Metal cations will generally not hydrolyze and, therefore, will be spectator ions that do not affect the pH.  Notable exceptions include Al³⁺ and Fe³⁺.

 

3. An anion will generally hydrolyze as a weak base:

 

A^- (aq) + H₂O (l) ⇌ HA (aq) + OH^- (aq)

 

We can determine K_b for the reaction above as follows:

 

A^- (aq) + H₂O (l) ⇌ HA (aq) + OH^- (aq)     K_b for A^- = ???

HA (aq) ⇌ H⁺ (aq) + A^- (aq)     K_a for HA = given in problem or found on K_a data table

_________________________________________________________________________________

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)     K_w = 1.0 x 10^-14

 

(K_b for A^-)(K_a for HA) = K_w

 

 

(so stronger acid HA = weaker conjugate base A^-)

 

4. Anions found in monoprotic strong acids (Cl^-, Br^-, I^-, NO₃^-, ClO₄^-) do not hydrolyze as weak bases to any significant extent, so these five anions will be spectator ions that do not affect the pH.

 

Using the guidelines above, we can predict if an ionic compound solution will be acidic, basic, or neutral as follows:

 

A. If both the cation and the anion are spectator ions, the solution will be neutral.

 

B. If only the cation hydrolyzes and the anion is a spectator ion, the solution will be acidic.

 

C. If the cation is a spectator ion and only the anion hydrolyzes, the solution will be basic.

 

D. If both the cation and anion hydrolyze, then we must compare the equilibrium constants for the two hydrolysis reactions:

 

if K_a for BH⁺ > K_b for A^-, then solution is acidic

if K_a for BH⁺ < K_b for A^-, then solution is basic

if K_a for BH⁺ = K_b for A^-, then solution is neutral

 

Sample Exercise 17L:

Predict whether a solution of each compound below will be acidic, basic, or neutral.  For solutions that are not neutral, show all relevant hydrolysis reactions that affect the pH and also calculate the equilibrium constant for each reaction you write using information from the following data tables:

 

 

 

a. Mg(NO₃)₂

b. CH₃NH₃Cl [composed of CH₃NH₃⁺ and Cl^-]

c. NaC₂H₃O₂

d. C₅H₅NHNO₂ [composed of C₅H₅NH⁺ and NO₂^-]

 

Solution:

a. Mg²⁺ and NO₃^- = spectator ions = solution is neutral

 

b. Cl^- = spectator ion, CH₃NH₃⁺ hydrolyzes as weak acid = solution is acidic:

 

CH₃NH₃⁺ (aq) ⇌ H⁺ (aq) + CH₃NH₂ (aq) 

 

 

c. Na⁺ = spectator ion, C₂H₃O₂^- hydrolyzes as a weak base = solution is basic:

 

C₂H₃O₂^- (aq) + H₂O (l) ⇌ HC₂H₃O₂ (aq) + OH^- (aq)

 

 

d. Both ions hydrolyze, so we must compare equilibrium constants:

 

C₅H₅NH⁺ (aq) ⇌ H⁺ (aq) + C₅H₅N (aq) 

 

 

NO₂^- (aq) + H₂O (l) ⇌ HNO₂ (aq) + OH^- (aq)

 

 

K_a > K_b = solution is acidic

 

 

To calculate the pH of a solution where only the cation hydrolyzes as a weak acid and the anion is a spectator ion, we complete a RICE chart and equilibrium calculation similar to that completed earlier for a neutral monoprotic weak acid.  To calculate the pH of a solution where only the anion hydrolyzes as a weak base and the cation is a spectator ion, we complete a RICE chart and equilibrium calculation similar to that completed earlier for a neutral weak base.

 

Sample Exercise 17M:

 

For each solution below, show any relevant hydrolysis reactions and calculate the pH.

 

a. 1.6 M NH₄I (K_b for NH₃ = 1.8 x 10^-5)

b. 0.84 M KClO (K_a for HClO = 3.0 x 10^-8)

 

Solution:

a. I^- = spectator ion, NH₄⁺ hydrolyzes as a weak acid:

 

 

 

 

b. K⁺ = spectator ion, ClO^- hydrolyzes as a weak base:

 

 

 

 

 

 

If a solution contains metal cations that are spectator ions along with anions of the type HA^-, a portion of the HA^- ions will ionize as weak acids while the remainder of the HA^- ions will hydrolyze as weak bases:

 

weak acid ionization reaction: 

HA^- (aq) ⇌ H⁺ (aq) + A^2- (aq)     K_a for HA^- = K_a2 for H₂A

weak base hydrolysis reaction:

HA^- (aq) + H₂O (l) ⇌ H₂A (aq) + OH^- (aq)     K_b for HA^- = K_w/(K_a1 for H₂A)

 

For anions of the type H₂A^-, the relevant reactions that affect the pH are:

 

weak acid ionization reaction: 

H₂A^- (aq) ⇌ H⁺ (aq) + HA^2- (aq)     K_a for H₂A^- = K_a2 for H₃A

weak base hydrolysis reaction:

H₂A^- (aq) + H₂O (l) ⇌ H₃A (aq) + OH^- (aq)     K_b for H₂A^- = K_w/(K_a1 for H₃A)

 

For anions of the type HA^2-, the relevant reactions that affect the pH are:

 

weak acid ionization reaction: 

HA^2- (aq) ⇌ H⁺ (aq) + A^3- (aq)     K_a for HA^2- = K_a3 for H₃A

weak base hydrolysis reaction:

HA^2- (aq) + H₂O (l) ⇌ H₂A^- (aq) + OH^- (aq)     K_b for HA^2- = K_w/(K_a for H₂A^- = K_a2 for H₃A)

 

To determine if an HA^-, H₂A^-, or HA^2- solution is acidic or basic, we must compare the equilibrium constants for the weak acid ionization reaction and the weak base hydrolysis reaction:

 

if K_a > K_b, then solution is acidic

if K_a < K_b, then solution is basic

 

Sample Exercise 17N:

Predict whether a solution of sodium bicarbonate, NaHCO₃, will be acidic or basic.  Show all relevant reactions that affect the pH and also give the value of the equilibrium constant for each reaction you write.  For carbonic acid, H₂CO₃, K_a1 = 4.3 x 10^-7 and K_a2 = 5.6 x 10^-11

 

Solution:

Na⁺ = spectator ion

 

weak acid ionization reaction: 

HCO₃^- (aq) ⇌ H⁺ (aq) + CO₃^2- (aq)    

K_a for HCO₃^- = K_a2 for H₂CO₃ = 5.6 x 10^-11

weak base hydrolysis reaction:

HCO₃^- (aq) + H₂O (l) ⇌ H₂CO₃ (aq) + OH^- (aq)    

K_b for HCO₃^- = 1.0 x 10^-14/(K_a1 for H₂CO₃ = 4.3 x 10^-7) = 2.3 x 10^-8

 

K_a < K_b = solution is basic

 

Section 17-9:  Reactions of Nonmetal Oxides with Water to Create Oxoacids

Recall from Chapter 10 that basic hydroxide ion is produced when a metal oxide reacts with water.  When a nonmetal oxide reacts with water, the product is an oxoacid.  Note that this is not a redox reaction, so the atoms do not undergo a change in oxidation number as the reaction proceeds.

 

nonmetal oxide + water → oxoacid with no change in oxidation numbers

 

For example, the reaction of SO₂ (oxidation number of S = +4) and water will produce H₂SO₃ (oxidation number of S = +4) rather than H₂SO₄ (oxidation number of S = +6).

 

Sample Exercise 17O:

 

 

Will the reaction of N₂O₅ and water produce HNO₂ or HNO₃?  Write a balanced equation for the reaction.

 

Solution:

 

The reaction of N₂O₅ (oxidation number of N = +5) and water will produce HNO₃ (oxidation number of N = +5) rather than HNO₂ (oxidation number of N = +3).  The balanced equation will be N₂O₅ + H₂O → 2 HNO₃.

 

Section 17-10:  Experiment – Determining the Molar Mass of an Unknown Monoprotic Weak Acid by Titration

Recall from Chapter 3 that we can determine the empirical formula of a compound if we know the percent composition by mass, after which we can determine the molecular formula using the experimentally determined molar mass. We can obtain the molar mass of an unknown monoprotic weak acid HA using titration as follows:

 

1. Dissolve a known mass of HA in water.

 

2. Titrate the HA solution with a metal hydroxide solution of known molarity from a buret.  Record the volume of metal hydroxide solution required to reach the equivalence point.

 

3.  Use stoichiometry from the net ionic equation HA (aq) + OH^- (aq) → H₂O (l) + A^- (aq) to calculate the moles of HA reacted, then divide the mass of HA by moles of HA to obtain the molar mass of HA.

   

Sample Exercise 17P:

a. An unknown monoprotic weak acid was found to be 54.53% carbon and 9.15% hydrogen by mass, with the remainder being oxygen.  Determine the empirical formula of the acid.

b. In a separate experiment, 6.7 grams of the acid was dissolved in 25 mL of water and then titrated with 0.85 M barium hydroxide.  The volume of base required to reach the equivalence point was 45 mL.  Calculate the molar mass and determine the molecular formula of the acid.

 

Solution:

 

a. 100% - 54.53% C - 9.15% H = 36.32%O by mass.

 

Assume one hundred grams of unknown acid:

 

 

4.540 mol C: 9.08 mol H: 2.270 mol O (divide each by 2.270)

= 2 mol C: 4 mol H: 1 mol O

Therefore, empirical formula is C₂H₄O.

 

b. 0.85 M Ba(OH)₂ → 0.85 M Ba²⁺ and 2 x 0.85 M = 1.7 M OH^-

Use stoichiometry from the net ionic equation HA (aq) + OH^- (aq) → H₂O (l) + A^- (aq) to calculate the moles of HA reacted, then divide the mass of HA by moles of HA to obtain the molar mass of HA.  Note that the volume of water used to dissolve the HA is not relevant in this calculation:

 

 

Since the acid is monoprotic, we can rewrite the molecular formula as HC₄H₇O₂.

 

 

 

Section 17-11:  pH of Buffer Solutions

One type of buffer solution contains significant concentrations of both a weak acid HA and its conjugate base A^-.  When HA and A^- are mixed in solution, the presence of a significant concentration of A^- reduces the ionization of HA in the reaction HA (aq) ⇌ H⁺ (aq) + A^- (aq).  This is an example of the common-ion effect.  We will generally assume that the common-ion effect reduces ionization of HA to an extent where the initial concentrations and moles of HA and A^- in the solution upon mixing are effectively equal to their equilibrium concentrations and moles.  Therefore, for the ionization of HA in an HA/A^- buffer solution, we can use initial rather than equilibrium concentrations and moles in the K_a expression below:

 

 

A variation of the expression above known as the Henderson-Hasselbalch equation can be used to calculate the pH of an HA/A^- buffer solution directly and can derived as follows:

 

 

Sample Exercise 17Q:

Calculate the pH of a solution containing 0.46 grams of NaF in 75 mL of 0.20 M HF (K_a = 6.8 x 10^-4).

 

Solution:

Na⁺ = spectator ion, HF/F^- = buffer solution:

 

 

An HA/A^- buffer solution is able to maintain a constant pH as water is added because the ratios [A^-]/[HA] and n_A^-/n_HA will not change upon dilution.  However, the solution will eventually cease functioning as an effective buffer if the volume is increased to a point where the concentrations of HA and A^- are too low.

 

 

A second type of buffer solution contains significant concentrations of both a weak base B and its conjugate acid BH⁺.  When B and BH⁺ are mixed in solution, the presence of a significant concentration of BH⁺ reduces the ionization of B in the reaction B (aq) + H₂O (l) ⇌ BH⁺ (aq) + OH^- (aq).  This is another example of the common-ion effect.  We will generally assume that the common-ion effect reduces ionization of B to an extent where the initial concentrations and moles of B and BH⁺ in the solution upon mixing are effectively equal to their equilibrium concentrations and moles.  Therefore, for the ionization of B in a B/BH⁺ buffer solution, we can use initial rather than equilibrium concentrations and moles in the K_b expression below:

 

 

A variation of the expression above can be used to calculate the pH of a B/BH⁺ buffer solution directly and can derived as follows:

 

 

Sample Exercise 17R:

Calculate the pH of a solution containing 3.0 grams of NH₄Cl in 30. mL of 1.6 M NH₃ (K_b = 1.8 x 10^-5).

 

Solution:

Cl^- = spectator ion, NH₃/NH₄⁺ = buffer solution:

 

A B/BH⁺ buffer solution is able to maintain a constant pH as water is added because the ratios [BH⁺]/[B] and n_BH⁺/n_B will not change upon dilution.  However, the solution will eventually cease functioning as an effective buffer if the volume is increased to a point where the concentrations of B and BH⁺ are too low.

 

 

Even if the amount of strong acid or strong base added to pure water is small, the pH can still change from pH = 7 by several units.  For example, if 0.001 mol H⁺ is added to water to create 1 L of 0.001 M H⁺, the new pH = -log (0.001) = 3.0.  Likewise, if 0.001 mol OH^- is added to water to create 1 L of 0.001 M OH^-, the new pOH = -log (0.001) = 3.0 and the new pH = 11.0.

 

On the other hand, the pH will only change slightly as small increments of strong acid or strong base are added to an HA/A^- or B/BH⁺ buffer solution.  The H⁺ or OH^- added will be neutralized by buffer components and the new pH of the buffer solution calculated as follows (based on 1:1 stoichiometric mole ratios and assuming the H⁺ or OH^- added is the limiting reagent):

 

OH^- added to HA/A^- buffer:

 

neutralization reaction:  HA (aq) + OH^- (aq) → A^- (aq) + H₂O (l)

pH = pK_a + log (n_A^- + n_OH^-)/(n_HA - n_OH^-)

(only valid if n_HA > n_OH^-)

 

H⁺ added to HA/A^- buffer:

 

neutralization reaction:  A^- (aq) + H⁺ (aq) → HA (aq)

pH = pK_a + log (n_A^- - n_H⁺)/(n_HA + n_H⁺)

(only valid if n_A^- > n_H⁺) 

 

H⁺ added to B/BH⁺ buffer:

 

neutralization reaction:  B (aq) + H⁺ (aq) → BH⁺ (aq)

pOH = pK_b + log (n_BH⁺ + n_H⁺) /(n_B - n_H⁺) = 14.00 - pH

(only valid if n_B > n_H⁺) 

 

OH^- added to B/BH⁺ buffer:

 

neutralization reaction:  BH⁺ (aq) + OH^- (aq) → B (aq) + H₂O (l)

pOH = pK_b + log (n_BH⁺ - n_OH^-)/(n_B + n_OH^-) = 14.00 - pH

(only valid if n_BH⁺ > n_OH^-) 

 

Sample Exercise 17S:

Write the net ionic equation for the neutralization reaction that occurs and calculate the pH when:

 

a. 0.001 mol HCl is added to the solution in Sample Exercise 17Q

b. 0.001 mol NaOH is added to the solution in Sample Exercise 17Q

c. 0.001 mol HCl is added to the solution in Sample Exercise 17R

d. 0.001 mol NaOH is added to the solution in Sample Exercise 17R

 

Solution:

a. Cl^- = spectator ion

neutralization reaction:  F^- (aq) + H⁺ (aq) → HF (aq)

pH = 3.17 + log (0.011 – 0.001) mol F^-/(0.015 + 0.001) mol HF = 2.97

 

b. Na⁺ = spectator ion

neutralization reaction:  HF (aq) + OH^- (aq) → F^- (aq) + H₂O (l)

pH = 3.17 + log (0.011 + 0.001) mol F^-/(0.015 - 0.001) mol HF = 3.10

 

c. Cl^- = spectator ion

neutralization reaction:  NH₃ (aq) + H⁺ (aq) → NH₄⁺ (aq)

pOH = 4.74 + log (0.056 + 0.001) mol NH₄⁺/(0.048 - 0.001) mol NH₃ = 4.82

pH = 14.00 – 4.82 = 9.18 

 

d. Na⁺ = spectator ion

neutralization reaction:  NH₄⁺ (aq) + OH^- (aq) → NH₃ (aq) + H₂O (l)

pOH = 4.74 + log (0.056 - 0.001) mol NH₄⁺/(0.048 + 0.001) mol NH₃ = 4.79

pH = 14.00 – 4.79 = 9.21

 

 

Rather than mixing HA and A^- or B and BH⁺ to create a buffer solution, we can also create an HA/A^- or B/BH⁺ buffer solution via a neutralization reaction if the H⁺ or OH^- added is the limiting reagent and the other reactant is in excess.  If that is not the case, then the method of calculating the pH of the resulting solution after the neutralization reaction is complete is also described below:

 

OH^- added to HA:

 

neutralization reaction:  HA (aq) + OH^- (aq) → A^- (aq) + H₂O (l)

n_HA > n_OH^-:  HA/A^- buffer created, pH = pK_a + log (n_A^- produced/n_HA excess)

n_HA = n_OH^-:  A^- solution created, pOH = 14.00 - pH calculated using RICE chart for hydrolysis of weak base A^-

n_HA < n_OH^-:  OH^-/A^- solution created, pOH = 14.00 – pH = -log [OH^-]_excess (hydrolysis of A^- negligible)

 

H⁺ added to A^-:

 

neutralization reaction:  A^- (aq) + H⁺ (aq) → HA (aq)

n_A^- > n_H⁺:  HA/A^- buffer created, pH = pKa + log (n_A^- excess/n_HA produced)

n_A^- = n_H⁺:  HA solution created, pH calculated using RICE chart for ionization of weak acid HA

n_A^- < n_H⁺:  H⁺/HA solution created, pH = -log [H⁺]_excess (ionization of HA negligible)

 

H⁺ added to B:

 

neutralization reaction:  B (aq) + H⁺ (aq) → BH⁺ (aq)

n_B > n_H⁺:  B/BH⁺ buffer created, pOH = pK_b + log (n_BH⁺ produced/n_B excess) = 14.00 - pH

n_B = n_H⁺:  BH⁺ solution created, pH calculated using RICE chart for hydrolysis of weak acid BH⁺

n_B < n_H⁺:  H⁺/BH⁺ solution created, pH = -log [H⁺]_excess (hydrolysis of BH⁺ negligible)

 

OH^- added to BH⁺:

 

neutralization reaction:  BH⁺ (aq) + OH^- (aq) → B (aq) + H₂O (l)

n_BH⁺> n_OH^-:  B/BH⁺ buffer created, pOH = pK_b + log (n_BH⁺ excess/n_B produced) = 14.00 - pH

n_BH⁺ = n_OH^-:  B solution created, pOH = 14.00 - pH calculated using RICE chart for ionization of weak base B

n_BH⁺< n_OH^-:  OH^-/B solution created, pOH = 14.00 – pH = -log [OH^-]_excess (ionization of B negligible)

 

 Sample Exercise 17T:

Write the net ionic equation for the neutralization reaction that occurs in each aqueous mixture below.  Which one of the four reactions creates a buffer solution?  For each mixture, describe the method of calculating the pH of the resulting solution after the neutralization reaction is complete.

 

a. 0.1 mol HBr + 0.1 mol CH₃NH₂ (a weak base) 

b. 0.3 mol HI + 0.1 mol KC₃H₅O₃

c. 0.3 mol NH₄NO₃ + 0.1 mol NaOH

d. 0.1 mol HNO₂ and 0.1 mol KOH

 

Solution:

a. Br^- = spectator ion

neutralization reaction:  CH₃NH₂ (aq) + H⁺ (aq) → CH₃NH₃⁺ (aq)

n of CH₃NH₂ = n_H⁺:  CH₃NH₃⁺ solution created, pH calculated using RICE chart for hydrolysis of weak acid CH₃NH₃⁺

 

b. I^- and K⁺ = spectator ions

neutralization reaction:  C₃H₅O₃^- (aq) + H⁺ (aq) → HC₃H₅O₃ (aq)

n of C₃H₅O₃^- < n_H⁺:  H⁺/ HC₃H₅O₃ solution created, pH = -log [H⁺]_excess (ionization of HC₃H₅O₃ negligible)

 

c. NO₃^- and Na⁺ = spectator ions

neutralization reaction:  NH₄⁺ (aq) + OH^- (aq) → NH₃ (aq) + H₂O (l)

n of NH₄⁺ > n_OH^-:  NH₃/ NH₄⁺ buffer created, pOH = pK_b + log (n of NH₄⁺ excess/n of NH₃ produced) = 14.00 - pH

 

d. K⁺ = spectator ion

neutralization reaction:  HNO₂ (aq) + OH^- (aq) → NO₂^- (aq) + H₂O (l)

n of HNO₂ = n_OH^-:  NO₂^- solution created, pOH = 14.00 - pH calculated using RICE chart for hydrolysis of weak base NO₂^-

 

 

  

 Section 17-12:  pH and Titration Curves

A titration curve is a graph showing the pH at different points during a titration:

 

Strong Acid + Strong Base Titration

 

When a strong acid is titrated with a strong base, the net ionic equation for the neutralization reaction is H⁺ (aq) + OH^- (aq) → H₂O (l), and a sketch of the titration curve is shown below:

 

 

Stoichiometry can be used to calculate the volume of base needed to reach the equivalence point as follows:

 

L_H⁺ × M_H⁺ = n_H⁺ = n_OH^-/ M_OH^- = L_OH^-

 

We can calculate the pH at different points on the titration curve above as follows:

 

Point A = initial:

pH = -log [H⁺]

 

Point B = between start of titration and equivalence point:

pH = -log [H⁺]_excess = -log [(n_H⁺ - n_OH^-)/L total]

 

Point C = equivalence point:

n_H+ = n_OH^- = only H₂O (l) present so pH = 7

 

Point D = after equivalence point:

pOH = -log [OH^-]_excess = -log [(n_OH^- - n_H⁺)/L total] = 14.00 - pH   

 

 Sample Exercise 17U:

a. Write the net ionic equation for the neutralization reaction that occurs during the titration of 24 mL of 0.30 M HCl with 0.20 M NaOH and calculate the volume of base needed to reach the equivalence point.

 

b. Calculate the following:

 

i. the initial pH

ii. the pH after 22 mL of base has been added

iii. the pH at the equivalence point

iv. the pH after 48 mL of base has been added

 

Solution:

a. Cl^- and Na⁺ = spectator ions, neutralization reaction:  H⁺ (aq) + OH^- (aq) → H₂O (l)

0.024 L (0.30 mol H⁺/1 L) = 0.0072 mol H⁺ = 0.0072 mol OH^- (1 L/0.20 mol OH^-) = 0.036 L = 36 mL base needed

 

b.

i. pH = -log (0.30) = 0.52

 

ii. 0.022 L (0.20 mol OH^-/1 L) = 0.0044 mol OH^- added

pH = -log [(0.0072 - 0.0044) mol H⁺ excess/(0.024 + 0.022) L total] = 1.22

 

iii. only H₂O present, pH = 7

 

iv. 0.048 L (0.20 mol OH^-/1 L) = 0.0096 mol OH^- added

pOH = - log [(0.0096 - 0.0072) mol OH^- excess/(0.024 + 0.048) L total] = 1.48

pH = 14.00 – 1.48 = 12.52

 

Strong Base + Strong Acid Titration

 

When a strong base is titrated with a strong acid, the net ionic equation for the neutralization reaction is OH^- (aq) + H⁺ (aq) → H₂O (l), and a sketch of the titration curve is shown below:

 

 

Stoichiometry can be used to calculate the volume of acid needed to reach the equivalence point as follows:

 

L_OH^- × M_OH^- = n_OH^- = n_H⁺/ M_H⁺ = L_H⁺

 

We can calculate the pH at different points on the titration curve above as follows:

 

Point A = initial:

pOH = -log [OH^-] = 14.00 - pH

 

Point B = between start of titration and equivalence point:

pOH = -log [OH^-]_excess = -log [(n_OH^- - n_H⁺)/L total] = 14.00 - pH

 

Point C = equivalence point:

n_OH^- = n_H⁺ = only H₂O (l) present so pH = 7

 

Point D = after equivalence point:

pH = -log [H⁺]_excess = -log [(n_H⁺ - n_OH^-)/L total]

 

We will demonstrate how to calculate the pH at different points in the titration of a strong base with a strong acid below in Practice Exercise 17-29.

 

 

Weak Acid + Strong Base Titration

 

When a weak acid is titrated with a strong base, the net ionic equation for the neutralization reaction is HA (aq) + OH^- (aq) → H₂O (l) + A^- (aq), and a sketch of the titration curve is shown below:

 

 

Stoichiometry can be used to calculate the volume of base needed to reach the equivalence point as follows:

 

L_HA × M_HA = n_HA = n_OH^-/ M_OH^- = L_OH^-

 

We can calculate the pH at different points on the titration curve above as follows:

 

Point A = initial pH calculated using RICE chart for ionization of weak acid HA:

 

 

 

Point B = between start of titration and equivalence point = HA/A^- buffer solution created:

n_OH^- added = n_HA reacted = n_A^- produced

n_HA initial - n_HA reacted = n_HA excess

pH = pK_a + log (n_A^- produced/n_HA excess)

halfway to equivalence point:  n_A^- produced = n_HA excess, so pH = pK_a

 

Point C = equivalence point, pOH = 14.00 - pH calculated using RICE chart for hydrolysis of weak base A^-:

 

 

pH > 7 at equivalence point

 

Point D = after equivalence point (hydrolysis of A^- negligible):

pOH = -log [OH^-]_excess = -log [(n_OH^- - n_HA)/L total] = 14.00 - pH   

 

Sample Exercise 17V:

a. Write the net ionic equation for the neutralization reaction that occurs during the titration of 16 mL of 0.20 M HC₂H₃O₂ (K_a = 1.8 x 10^-5) with 0.10 M KOH and calculate the volume of base needed to reach the equivalence point.

 

b. Calculate the following:

 

i. the initial pH

ii. the pH after 14 mL of base has been added

iii. the pH at the equivalence point

iv. the pH after 44 mL of base has been added

 

Solution:

a. K⁺ = spectator ion, neutralization reaction:  HC₂H₃O₂ (aq) + OH^- (aq) → H₂O (l) + C₂H₃O₂^- (aq)

0.016 L (0.20 mol HC₂H₃O₂/1 L) = 0.0032 mol HC₂H₃O₂ = 0.0032 mol OH^- (1 L/0.10 mol OH^-) = 0.032 L = 32 mL base needed

 

b.

i.

 

 

ii.

0.014 L (0.10 mol OH^-/1 L) = 0.0014 mol OH^- added = 0.0014 mol HC₂H₃O₂ reacted = 0.0014 mol C₂H₃O₂^- produced

0.0032 mol HC₂H₃O₂ initial - 0.0014 mol HC₂H₃O₂ reacted = 0.0018 mol HC₂H₃O₂ excess

pK_a for HC₂H₃O₂  = -log (1.8 x 10^-5) = 4.74

pH = 4.74 + log (0.0014 mol C₂H₃O₂^-/0.0018 mol HC₂H₃O₂) = 4.63

 

iii.

 

 

iv. 0.044 L (0.10 mol OH^-/1 L) = 0.0044 mol OH^- added

pOH = - log [(0.0044 - 0.0032) mol OH^- excess/(0.016 + 0.044) L total] = 1.70

pH = 14.00 – 1.70 = 12.30

 

 

Weak Base + Strong Acid Titration

 

When a weak base is titrated with a strong acid, the net ionic equation for the neutralization reaction is B (aq) + H⁺ (aq) → BH⁺ (aq), and a sketch of the titration curve is shown below:

 

 

Stoichiometry can be used to calculate the volume of acid needed to reach the equivalence point as follows:

 

L_B × M_B = n_B = n_H⁺/ M_H⁺ = L_H⁺

 

We can calculate the pH at different points on the titration curve above as follows:

 

Point A = initial pH calculated using RICE chart for ionization of weak base B:

 

 

 

Point B = between start of titration and equivalence point = B/BH⁺ buffer solution created:

n_H⁺ added = n_B reacted = n_BH⁺ produced

n_B initial - n_B reacted = n_B excess

pOH = pK_b + log (n_BH⁺ produced/n_B excess) = 14.00 - pH

halfway to equivalence point:  n_BH⁺ produced = n_B excess, so pOH = pK_b = 14.00 - pH

 

Point C = equivalence point, pH calculated using RICE chart for hydrolysis of weak acid BH⁺:

 

 

pH < 7 at equivalence point

 

Point D = after equivalence point (hydrolysis of BH⁺ negligible):

pH = -log [H⁺]_excess = -log [(n_H⁺ - n_B)/L total]

 

Sample Exercise 17W:

a. Write the net ionic equation for the neutralization reaction that occurs during the titration of 54 mL of 0.10 M NH₃ (K_b = 1.8 x 10^-5) with 0.30 M HCl and calculate the volume of acid needed to reach the equivalence point.

 

b. Calculate the following:

 

i. the initial pH

ii. the pH after 11 mL of acid has been added

iii. the pH at the equivalence point

iv. the pH after 32 mL of acid has been added

 

Solution:

a. Cl^- = spectator ion, neutralization reaction:  NH₃ (aq) + H⁺ (aq) → NH₄⁺ (aq)

0.054 L (0.10 mol NH₃/1 L) = 0.0054 mol NH₃ = 0.0054 mol H⁺ (1 L/0.30 mol H⁺) = 0.018 L = 18 mL acid needed

 

b.

i.

 

 

ii.

0.011 L (0.30 mol H⁺/1 L) = 0.0033 mol H⁺ added = 0.0033 mol NH₃ reacted = 0.0033 mol NH₄⁺ produced

0.0054 mol NH₃ initial - 0.0033 mol NH₃ reacted = 0.0021 mol NH₃ excess

pK_b for NH₃  = -log (1.8 x 10^-5) = 4.74

pOH = 4.74 + log (0.0033 mol NH₄⁺/0.0021 mol NH₃) = 4.94

pH = 14.00 – 4.94 = 9.06

 

iii.

 

 

iv. 0.032 L (0.30 mol H⁺/1 L) = 0.0096 mol H⁺ added

pH = - log [(0.0096 - 0.0054) mol H⁺ excess/(0.054 + 0.032) L total] = 1.31

 

 

 

Section 17-13:  Acid-Base Indicators

One type of acid-base indicator is a weak acid where the neutral acid HInd has a different color than its conjugate base Ind^-:

 

HInd (= color 1) ⇌ H⁺ + Ind^- (= color 2)

 

The pH of the solution being tested will determine the ratio of moles of Ind^- to moles of HInd and, therefore, the color of the indicator as follows:

 

pH of solution = pK_a for HInd + log (n_Ind^-/n_HInd)

 

pH << pK_a:  n_Ind^- <<  n_HInd = color 1

pH = pK_a:  n_Ind^- = n_HInd = mixture of color 1 and color 2

pH >> pK_a:  n_Ind^-  >>  n_HInd = color 2

 

During each titration discussed above, the pH changes sharply by several units near the equivalence point.  An appropriate indicator for a titration will have one color before the equivalence point and then instantaneously change to a different color after the equivalence point due to the sharp change in pH.  The indicator chosen should have pK_a for HInd close to the pH at the equivalence point:

 

titration:  choose indicator with pK_a for HInd close to pH at equivalence point

 

Sample Exercise 17X:

Which indicator, thymolphthalein (K_a = 1 x 10^-10) or bromcresol green (K_a = 8 x 10^-6), would be the better choice for the titration in Sample Exercise 17W?

 

Solution:

pH at equivalence point = 5.19

pK_a for thymolphthalein = -log (1 x 10^-10) = 10.0

pK_a for bromcresol green = -log (8 x 10^-6) = 5.1

pK_a for bromcresol green close to pH at equivalence point, so bromcresol green is better choice

 

 

 

Chapter 17 Practice Exercises and Review Quizzes:

 

 

 

 

 

 

 

 

 

 

 

 

 

    

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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